1. ## Order of element

How do you prove that the order of a in Zn (where n is an integer greater than or equal to 1) viewed as a group of order n with respect to addition is
n/(gcd(a,n))?

2. Originally Posted by PvtBillPilgrim
How do you prove that the order of a in Zn (where n is an integer greater than or equal to 1) viewed as a group of order n with respect to addition is
n/(gcd(a,n))?
You need to show that:

1) $a^{n/\gcd(a,n)}=0$

2)It is the smallest positive exponent which makes this statement true.

3. Do you mind elaborating on this?

By the way, I'm in a university abstract algebra course. He introduces some group theory at the end. Thanks for the help anyway.

4. Originally Posted by PvtBillPilgrim
Do you mind elaborating on this?

By the way, I'm in a university abstract algebra course. He introduces some group theory at the end. Thanks for the help anyway.
Let $G$ be a cyclic group (finite). And let $|G|=n$. It means the $G$ has a generator that is $=G$ for some $a\in G$.

Let $b\in G$ thus, $b=a^c$ for $1\leq c\leq n$.

We need to find the smallest $m$ such as,
$b^m=(a^c)^m=a^{cm}=e$
By the properties of cyclic groups and the fact that $a$ is a generator it is equivalent to saying $n$ divides $cm$.
Thus, we need the smallest $m$ such that,
$\frac{cm}{n}$ is an integer.
We can write it as, (by dividing by $d=\gcd (c,n)$
$\frac{m(c/d)}{(n/d)}$
But,
$\gcd \left( c/d,n/d \right)=1$
Thus,
$n/d$ divides $m$.
The smallest such $m$ is of course,
$n/d$

Thus, the order of any element is,
$\frac{n}{\gcd(c,n)}$